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Magazine Chapter 5: Problem 185
In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2 . \(\int_{0}^{\pi / 4} \sec \theta \tan \theta\)
Short Answer
Expert verified
The value of the integral is \(\sqrt{2} - 1\).
Step by step solution
01
Identify the Integral to Evaluate
We have the integral \( \int_{0}^{\pi / 4} \sec \theta \tan \theta \, d\theta \). We need to evaluate this using the Fundamental Theorem of Calculus, Part 2.
02
Find the Antiderivative
The antiderivative of \( \sec \theta \tan \theta \) is \( \sec \theta \), since the derivative of \( \sec \theta \) is \( \sec \theta \tan \theta \). Thus, \( F(\theta) = \sec \theta \).
03
Apply the Fundamental Theorem of Calculus, Part 2
According to the Fundamental Theorem of Calculus, Part 2, if \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). Here, \( F(\theta) = \sec \theta \), \( a = 0 \), and \( b = \pi / 4 \).
04
Evaluate the Antiderivative at the Upper Limit
Evaluate \( F(\theta) \) at \( \theta = \pi / 4 \): \( F(\pi / 4) = \sec(\pi / 4) = \frac{1}{\cos(\pi / 4)} = \sqrt{2} \).
05
Evaluate the Antiderivative at the Lower Limit
Evaluate \( F(\theta) \) at \( \theta = 0 \): \( F(0) = \sec(0) = \frac{1}{\cos(0)} = 1 \).
06
Calculate the Definite Integral Value
Compute \( F(\pi / 4) - F(0) = \sqrt{2} - 1 \).
07
Conclude the Solution
Therefore, the value of the integral \( \int_{0}^{\pi / 4} \sec \theta \tan \theta \, d\theta \) is \( \sqrt{2} - 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
The concept of a definite integral revolves around finding the total accumulation of quantities, often represented as the area under a curve in calculus. In our exercise, the definite integral \( \int_{0}^{\pi / 4} \sec \theta \tan \theta \, d\theta \) is to be evaluated. The definite integral has two limits of integration: the lower limit \( \theta = 0 \) and the upper limit \( \theta = \pi / 4 \). These limits are crucial as they define the interval over which the integration is performed.
The result of a definite integral depends not just on the antiderivative of the function, but also on these limits. By applying the Fundamental Theorem of Calculus, one can compute the difference between the values of the antiderivative evaluated at these two points. This theorem simplifies the computation, allowing us to convert an integration problem into a subtraction problem of two derived values.
The result of a definite integral depends not just on the antiderivative of the function, but also on these limits. By applying the Fundamental Theorem of Calculus, one can compute the difference between the values of the antiderivative evaluated at these two points. This theorem simplifies the computation, allowing us to convert an integration problem into a subtraction problem of two derived values.
Antiderivative
An antiderivative of a function is another function whose derivative is the original function. For example, in our problem, we needed to find the antiderivative of \( \sec \theta \tan \theta \), which turns out to be \( \sec \theta \). This is because the derivative of \( \sec \theta \) yields \( \sec \theta \tan \theta \).
In calculus, the process of finding an antiderivative is called integration. Finding antiderivatives can be thought of as reversing differentiation. Once we have the antiderivative, it forms the basis for calculating the definite integral using the values at the specified limits of integration. Through this process, the complex task of evaluating a definite integral is streamlined into simple arithmetic using the antiderivative.
In calculus, the process of finding an antiderivative is called integration. Finding antiderivatives can be thought of as reversing differentiation. Once we have the antiderivative, it forms the basis for calculating the definite integral using the values at the specified limits of integration. Through this process, the complex task of evaluating a definite integral is streamlined into simple arithmetic using the antiderivative.
Trigonometric Functions
Trigonometric functions are essential in calculus, particularly when dealing with periodic phenomena. In this exercise, the focus was on the functions \( \sec \theta \) and \( \tan \theta \). The secant function \( \sec \theta \) is essentially the reciprocal of the cosine function, while the tangent function \( \tan \theta \) is the ratio of the sine function over the cosine function.
Understanding the properties and derivatives of trigonometric functions is vital, as they frequently appear in problems involving geometry and oscillation. These functions can behave in more complex manners than polynomial functions, making them interesting and occasionally challenging to work with. Recognizing their derivatives is key to applying them effectively in calculus, as seen in our problem where the derivative relations were crucial for identifying the antiderivative and calculating the integral.
Understanding the properties and derivatives of trigonometric functions is vital, as they frequently appear in problems involving geometry and oscillation. These functions can behave in more complex manners than polynomial functions, making them interesting and occasionally challenging to work with. Recognizing their derivatives is key to applying them effectively in calculus, as seen in our problem where the derivative relations were crucial for identifying the antiderivative and calculating the integral.
